Uva 11401 - Triangle Counting 解题报告（计数）

Problem G
Triangle Counting

Input: Standard Input

Output: Standard Output

 

You are given n rods of length 1, 2…, n. You have to pick any 3 of them & build a triangle. How many distinct triangles can you make? Note that, two triangles will be considered different if they have at least 1 pair of arms with different length.

 

Input

 

The input for each case will have only a single positive integer n (3<=n<=1000000). The end of input will be indicated by a case with n<3. This case should not be processed.

 

Output

 

For each test case, print the number of distinct triangles you can make.

 

Sample Input                                                  Output for Sample Input

5

8

0

3

22

 

    解题报告： 分析。假设现在最场边的长度为n。另外两条边的长度为a和b。由三角形的性质可以知道n-a<b<n。我们递推。如果a=1，那么该不等式无解；如果a=2，该不等式有1个解；直到a=n-2，不等式有n-2个解。那么一共有(n-2)(n-1)/2个解。当然，两条边除了相同相同长度，其他长度对都出现了两次，需要去重。相同长度的三角形数量为(n-1)/2。n长度以内的三角形是为最长边为3-n所有的三角形，所以预处理求和。代码如下：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long LL;const int maxn = 1000010;LL num[maxn];void init(){    for(int i=3;i<maxn;i++)        num[i]=num[i-1]+((long long)(i-1)*(i-2)/2-(i-1)/2)/2;}int main(){    init();    int n;    while(~scanf("%d",&n) && n>=3)        printf("%lld\n", num[n]);}